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If f(x)= \int_{0}^{x}t (\sin x-\sin t)dt

then 

  • Option 1)

    f'''(x)+ f''(x)= \sin x

  • Option 2)

    f'''(x)+ f''(x)- f'(x)= \cos x

  • Option 3)

    f'''(x)+ f'(x)= \cos x - 2x \sin x

  • Option 4)

    f'''(x)- f''(x)= \cos x- 2x \sin x

 

Answers (2)

best_answer

As we have learned

NEWTON LEIBNITZ THEOREM -

\frac {d}{dt}\left ( \int_{f(t)}^{\phi (t))}F(x)dx \right )=F(\phi(t))\phi^{'}(t)-F(f(t))f^{'}(t)

-

 

 f(x)= \int_{0}^{x}t\sin x dt -\int_{0}^{x}t \sin t dt

=\frac{x^{2}}{2}\sin x - \int_{0}^{x}t \sin t dt

f'(x)= x\sin x + \frac{x^{2}}{2} \cos x - x\sin x

f'(x)= \frac{x^{2}}{2} \cos x

f''(x)= {x} \cos x - \frac{x^{2}}{2}\sin x

f'''(x)= -{x} \sin x + \cos x - x\sin x- \frac{x^{2}}{2}\cos x

Thus f'''(x)+f'(x)= \cos x-2x\sin x

 

 


Option 1)

f'''(x)+ f''(x)= \sin x

This is incorrect

Option 2)

f'''(x)+ f''(x)- f'(x)= \cos x

This is incorrect

Option 3)

f'''(x)+ f'(x)= \cos x - 2x \sin x

This is correct

Option 4)

f'''(x)- f''(x)= \cos x- 2x \sin x

This is incorrect

Posted by

Himanshu

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